Calculate cell potential at non-standard conditions.

Nernst Equation — E = E° - (RT/nF)ln Q

Question

A Daniell cell operates at 25°C. Given:

E°_\text{cell} = +1.10 \text{ V}

[\text{Zn}^{2+}] = 0.1 \text{ M}, \quad [\text{Cu}^{2+}] = 0.01 \text{ M}

Calculate the cell potential $E_\text{cell}$ at these non-standard conditions.

The cell reaction: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

Solution — Step by Step

E_\text{cell} = E°_\text{cell} - \frac{RT}{nF} \ln Q

At 25°C (298 K), the factor $\frac{RT}{F} = 0.02569$ V. Using $\log_{10}$ instead of $\ln$ , this simplifies to the form you’ll use in 90% of problems:

E_\text{cell} = E°_\text{cell} - \frac{0.0591}{n} \log Q

Look at the half-reactions:

\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad \text{(oxidation at anode)}

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad \text{(reduction at cathode)}

Both half-reactions involve 2 electrons, so $n = 2$ .

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.1}{0.01} = 10

We put products in the numerator and reactants in the denominator. Pure solids (Zn, Cu) don’t appear in Q — their activity is 1.

E_\text{cell} = 1.10 - \frac{0.0591}{2} \log(10)

E_\text{cell} = 1.10 - \frac{0.0591}{2} \times 1

E_\text{cell} = 1.10 - 0.02955

\boxed{E_\text{cell} \approx 1.07 \text{ V}}

Why This Works

The standard cell potential $E°$ assumes all species are at 1 M concentration and 1 atm pressure — conditions that almost never exist in a real cell. The Nernst equation corrects for this by accounting for the actual concentrations through the reaction quotient $Q$ .

When $Q > 1$ , products are already accumulating relative to reactants. The system has less “driving force” left, so $E$ drops below $E°$ . Here, $[\text{Zn}^{2+}]$ is ten times greater than $[\text{Cu}^{2+}]$ , so the cell is partially “done” — and we see the potential fall from 1.10 V to 1.07 V.

When $Q = 1$ (all concentrations at 1 M), $\log Q = 0$ and the equation correctly gives $E = E°$ . This is the internal logic check that proves the formula is consistent.

Alternative Method — Using the ln form directly

Some questions in JEE give data with $R$ , $T$ , $F$ explicitly. Use the full form:

E_\text{cell} = 1.10 - \frac{(8.314)(298)}{(2)(96500)} \ln(10)

= 1.10 - (0.01285)(2.303)

= 1.10 - 0.02959 \approx 1.07 \text{ V}

Same answer, different route. The $0.0591/n$ shortcut is derived exactly from this — memorise the shortcut, but understand where it comes from.

The factor $0.0591$ is only valid at 25°C (298 K). If a question specifies a different temperature — say 37°C or 50°C — you must use the full $RT/nF$ form. JEE 2023 had a variant at a different temperature specifically to catch students using 0.0591 blindly.

Common Mistake

Writing Q upside down. Many students write $Q = [\text{Cu}^{2+}]/[\text{Zn}^{2+}]$ — putting the reactant ion in the numerator because “it’s on the left side of the cell notation.” Always write Q as products over reactants for the overall cell reaction, not the half-reaction. Here, $\text{Zn}^{2+}$ is the product and $\text{Cu}^{2+}$ is the reactant, so $Q = 0.1/0.01 = 10$ . Flipping this gives $Q = 0.1$ , which makes $\log Q$ negative, incorrectly increasing the cell potential above $E°$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the ln form directly

Common Mistake

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